Here is the statement of the wikipedia page for mathematical symbols and i'm struggling to understand it in context;

Statement: Every finite, non-empty, ordered set has a largest element. Otherwise, let's assume that X is a finite, non-empty, ordered set with no largest element. Then, for some $x_1$ is a member of $X$, there exists an $x_2$ in $X$ with $x_1 < x_2$, but then there's also an $x_3$ in $X$ with $x_2 < x_3$, and so on. Thus, $x_1, x_2, x_3, ...$ are distinct elements in X. ↯ X is finite.

My thinking:

Every finite, non-empty, order set has a largest element. (True, a non-empty set has a element which has a quantity.)

Otherwise X is finite, non-empty, ordered with no largest element. (This can't happen, so it's unnecessary to make this statement?)

Then, for some $x_1$ in $X$ there exists an $x_2$ in $X$ with $x_1 < x_2$ (Well maybe not? if X is of length 1; Is this the contradictory statement?)

It then goes on to say this holds true for all $x_{n+1}$ (If what was stated before then this can't hold?)

My question is what does this ↯ mean in this statement, at what point is the contradiction? Isn't that kind of ambiguous in a mathematical proof?

Can someone shed some light, i'm getting all kinds of confused.


You should read this as "Contradiction, because $X$ is finite."

The assumption is that X is finite and has no largest element, and argument proceeds to show why that doesn't work. (Since it's finite, the chain $x_1 < x_2 < x_3 < ...$ must stop at some $x_n$)

  • $\begingroup$ That makes so much more sense.... Thanks... Feel like a bit of an idiot haha $\endgroup$ – Alex White Dec 16 '17 at 10:02
  • $\begingroup$ @AlexWhite Always ask questions. :) $\endgroup$ – naslundx Dec 16 '17 at 10:09

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